Find all critical points of f(z)=(3z^2-6)^2(1-5z)^4

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Find all critical points of f(z)=(3z^2-6)^2(1-5z)^4

Mathematics
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z = 2 and z = 1/5
Take the derivative of f(z) with respect to z and set it equal to zero. The values of z that result will be critical points; don't know if you want to distinguish between different types though.
I understand how to find them - I was having problems with setting the derivative to zero.

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Alright; give me a minute and I'll try it out. :)
I've got f'(z)=6(z^2-2)(6z)(1-5z)^4-15(z^2-2)4(1-5z)^3 for the derivative. Look like what you got?
When you differentiate, it should come out to this:\[f'(z) = 6*(1-5z)^4*(3z^2-6)*2z-20*(1-5z)^3*(3x^2-6)^2\] Remember to not get sidetracked by careless errors; they're very prevalent in this problem.
Sorry, accidentally stuck an x into the last factor when it was meant to be a z :P
The image cuts off at (3x^2 - 6)...is that the end? Just making sure
I get a pretty simple quadratic equation in the end..yep.
Hmm..How did you get to the quadratic equation? Sorry I don't understand
I just moved the entire right term (starting with -20*(1-5z)^3...) to the left hand side of the equation, and started simplifying. Things begin cancelling immediately and it breaks down into a quadratic equation (or so I believe)
You set them equal to each other and simplify? How can they simplify if they are being multiplied to each other - I though you can't do that.
No...I took that derivative and set it equal to zero, then when I moved the term beginning with -20 to the opposite side (where the 0 would be) I began cancelling common factors.
Did you get 0 = -60z^2 +12z - 20 at some point?
Did you get 0 = -60z^2 +12z - 20 at some point?
Hm...let me show you what I did: \[0 = 6*(1-5z)^4*(3z^2-6)* (2z) - 20 * (1-5z)^3*(3z^2-6)^2\]\[20 * (1-5z)^3*(3z^2-6)^2= 6*(1-5z)^4*(3z^2-6)* (2z)\](cancel like terms and move coefficients)\[\frac{5}{3}*(3z^2-6) = (1-5z)*z\]\[5z^2-10=z-5z^2\]\[10z^2-z-10 = 0\] and you can solve after that.
Ohhhhh...Okay. Thank you very much! I appreciate your help.
Glad to help. :)
The first person to answer this was incorrect, right? You should get some weird decimal numbers after using quadratic formula?

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